ELE-470 · Lesson 4 of 12 · 32 min
Magnetics and capacitors
Semiconductors switch power, but inductors, transformers and capacitors store and transfer it. In most converters these passive components take up more volume, weigh more and cost more than the switches, and they often set the limits on efficiency, power density and lifetime. They are also the components that engineers most often design from first principles, because standard parts rarely match a converter's exact needs. Lesson 1 described their role briefly. This lesson covers the physics of magnetic components, core materials, the area-product method for designing an inductor, core and winding losses and how they change with frequency, transformer design for switching converters, and the selection of capacitors, including the ripple current, voltage derating and lifetime issues that cause many field failures.
Magnetic fundamentals
A current \(i\) flowing through \(N\) turns around a magnetic path of length \(l\) produces a magnetic field strength \(H\) given by Ampère's law, \(N i = H l\). The field produces a flux density \(B = \mu H\) in the core, where the permeability \(\mu = \mu_0 \mu_r\) is the product of the permeability of free space and the relative permeability of the material, several thousand for ferrite. The flux \(\Phi = B A_e\) through the core's cross-sectional area links the turns, and Faraday's law relates the voltage across the winding to its rate of change, \(v = N \, d\Phi/dt\).
It is convenient to describe a magnetic path by its reluctance, analogous to resistance in an electric circuit:
Reluctances in series add. Inductors for power converters almost always include an air gap in series with the core: because air has a relative permeability of 1, even a gap of a millimeter has far more reluctance than the rest of a ferrite core, so the inductance is set by the gap, which is easy to control, rather than by the core's permeability, which varies with temperature and from part to part. The gap also stores most of the energy, \(\tfrac{1}{2} L i^2\), and it allows the core to carry a large DC current without saturating.
Saturation is the limit that matters most. As \(B\) approaches the material's saturation flux density, the permeability collapses, inductance falls sharply, and current rises rapidly, which can destroy switches in microseconds. Designs therefore keep the peak flux density, at the peak current and the highest operating temperature, safely below saturation.
Core materials
| Material | Saturation flux density | Typical frequency range | Typical uses |
|---|---|---|---|
| Silicon steel laminations | About 1.5 to 1.8 T | 50/60 Hz to a few kHz | Line-frequency transformers, chokes, motors |
| Amorphous and nanocrystalline alloys | About 1.2 to 1.6 T | Up to tens of kHz | Medium-frequency transformers, common-mode chokes, efficient distribution transformers |
| Powder cores (iron, sendust, MPP) | About 0.8 to 1.5 T | Up to a few hundred kHz | Filter inductors, PFC inductors with DC bias |
| Manganese-zinc ferrite | About 0.4 to 0.5 T at 25 °C, lower when hot | 20 kHz to about 1 MHz | Switching transformers and inductors |
| Nickel-zinc ferrite | About 0.3 T | Above 1 MHz | High-frequency and EMI components |
Ferrites have high resistivity, so eddy currents inside them are small even at hundreds of kilohertz, but their low saturation flux density and its fall with temperature (to about 0.35 T at 100 °C for many grades) limit how much energy a given core can handle. Powder cores distribute a tiny air gap throughout the material, between insulated metal particles, so they need no discrete gap and saturate gradually: their inductance falls steadily as DC current rises, which designers must account for. Laminated and tape-wound metal cores carry high flux but suffer eddy current losses that rise rapidly with frequency.
Designing an inductor
Two constraints fix the size of an inductor. The core must carry the peak flux without saturating, and the window (the space for the winding) must hold enough copper to carry the RMS current at an acceptable current density. Combining them gives the area product, the product of the core's effective area \(A_e\) and window area \(A_w\), that a design requires:
where \(B_{max}\) is the allowed peak flux density, \(J\) the current density in the copper (commonly 3 to 6 A/mm²) and \(K_u\) the window utilization, the fraction of the window actually filled with copper after allowing for insulation, bobbins and air (often 0.3 to 0.5). The designer picks the smallest standard core with enough area product, then calculates the turns from \(N = L I_{pk} / (B_{max} A_e)\) and the gap length from \(l_g \approx \mu_0 N^2 A_e / L\).
import math
# Designing a gapped ferrite inductor for a buck converter by the area-product method
L = 100e-6 # H
I_pk, I_rms = 32.0, 28.0
B_max = 0.30 # T, below ferrite saturation with margin
J = 5e6 # A/m2 current density (5 A/mm2)
K_u = 0.4 # fraction of the window filled with copper
mu0 = 4e-7 * math.pi
Ap_need = L * I_pk * I_rms / (B_max * J * K_u) # m4
print(f"required area product: {Ap_need * 1e12:,.0f} mm4")
cores = { # approximate data: Ae (mm2), Aw (mm2), mean turn length (mm)
"E32/16/9": (83, 99, 66),
"E42/21/15": (178, 256, 93), "E42/21/20": (233, 256, 99),
"E55/28/21": (353, 370, 116), "E65/32/27": (535, 540, 150),
}
for name, (Ae, Aw, mlt) in cores.items():
Ap = Ae * Aw
ok = "OK" if Ap * 1e-12 >= Ap_need else "too small"
print(f"{name}: Ap = {Ap:7,.0f} mm4 {ok}")
name = next(k for k, v in cores.items() if v[0] * v[1] * 1e-12 >= Ap_need) # smallest core that fits
Ae, Aw, mlt = cores[name][0] * 1e-6, cores[name][1] * 1e-6, cores[name][2]
N = math.ceil(L * I_pk / (B_max * Ae))
gap = mu0 * N ** 2 * Ae / L # ignoring core reluctance and fringing
A_cu = I_rms / J
fill = N * A_cu / Aw
R_dc = 1.72e-8 * N * mlt * 1e-3 / A_cu * 1.3 # copper resistivity at about 100 C (x1.3)
print(f"chosen {name}: N = {N} turns, air gap {gap * 1e3:.2f} mm, copper area {A_cu * 1e6:.1f} mm2 per turn")
print(f"window fill {fill:.0%}, DC resistance {R_dc * 1e3:.1f} milliohm, copper loss {I_rms ** 2 * R_dc:.1f} W")
print(f"check flux at peak current: {L * I_pk / (N * Ae):.3f} T")required area product: 149,333 mm4 E32/16/9: Ap = 8,217 mm4 too small E42/21/15: Ap = 45,568 mm4 too small E42/21/20: Ap = 59,648 mm4 too small E55/28/21: Ap = 130,610 mm4 too small E65/32/27: Ap = 288,900 mm4 OK chosen E65/32/27: N = 20 turns, air gap 2.69 mm, copper area 5.6 mm2 per turn window fill 21%, DC resistance 12.0 milliohm, copper loss 9.4 W check flux at peak current: 0.299 T
The smallest suitable core is an E65 size with 20 turns and a gap of about 2.7 mm. A gap this large produces significant fringing flux, which spreads out around the gap, increases the effective gap area (so the actual inductance is somewhat higher than the simple formula predicts) and induces eddy currents in nearby winding turns. Designers reduce these effects by splitting the gap into several smaller gaps along the center leg, keeping windings away from the gap, or using powder cores with distributed gaps. Final values are always checked by measurement.
Core and winding losses
Core losses come from hysteresis, the energy needed to reverse the magnetic domains in each cycle, and eddy currents induced in the core material. For sinusoidal excitation, measured losses per unit volume are fitted by the Steinmetz equation:
where \(f\) is the frequency, \(B_{pk}\) the peak AC flux density (half the peak-to-peak swing), and \(k\), \(\alpha\) (typically 1.1 to 1.8) and \(\beta\) (typically 2.3 to 3) are fitted to the manufacturer's data for each material and temperature. Converter waveforms are not sinusoidal, so improved versions such as the improved generalized Steinmetz equation (iGSE) calculate losses from the rate of change of flux. Core losses depend on the AC part of the flux only; the DC bias in an inductor adds little, although it affects permeability in powder cores.
Winding losses start with the DC resistance of the copper, \(I^2 R\), which rises by about 0.4 percent per kelvin. At high frequency, two further effects increase resistance. The skin effect confines AC current to a layer near the surface of a conductor, with a depth that shrinks with the square root of frequency:
about 0.24 mm in copper at 100 kHz and 100 °C. The proximity effect, in which the field from neighboring turns and layers induces eddy currents in each conductor, is often worse than the skin effect in multilayer windings, and its losses grow rapidly with the number of layers. Remedies include litz wire (many fine insulated strands twisted so that each occupies every position in the bundle), thin copper foil, fewer layers, and interleaving primary and secondary windings in transformers.
import math
# Core loss (Steinmetz equation) and skin effect as switching frequency rises
k, a, b = 2.4e-3, 1.4, 2.6 # ferrite at about 100 C: Pv in mW/cm3 with f in Hz and B in T (illustrative)
def core_loss(f, B_peak):
return k * f ** a * B_peak ** b
V_core = 79.0 # cm3, about an E65 core
rho = 2.2e-8 # copper resistivity at about 100 C, ohm m
mu0 = 4e-7 * math.pi
d_wire = 2.0e-3 # solid round wire diameter, m
print(" f (kHz) ripple B (T) core loss (W) skin depth (mm) Rac/Rdc 2 mm wire")
for f in (25e3, 50e3, 100e3, 200e3, 400e3):
B = 0.05 * (100e3 / f) # same inductor: ripple current, and so ripple flux, fall as 1/f
P = core_loss(f, B) * V_core / 1000
delta = math.sqrt(rho / (math.pi * f * mu0))
ratio = 1.0 if d_wire < 2 * delta else d_wire / (4 * delta) + 0.25
print(f"{f / 1e3:8.0f} {B:12.4f} {P:13.2f} {delta * 1e3:15.3f} {ratio:17.2f}")
# Strands needed in litz wire at 100 kHz to keep each strand below one skin depth
f = 100e3
delta = math.sqrt(rho / (math.pi * f * mu0))
A_cu = 5.6e-6
strand = delta # strand diameter equal to the skin depth
n = math.ceil(A_cu / (math.pi * strand ** 2 / 4))
print(f"litz at 100 kHz: strands of {strand * 1e3:.2f} mm, about {n} strands for {A_cu * 1e6:.1f} mm2") f (kHz) ripple B (T) core loss (W) skin depth (mm) Rac/Rdc 2 mm wire
25 0.2000 4.15 0.472 1.31
50 0.1000 1.80 0.334 1.75
100 0.0500 0.79 0.236 2.37
200 0.0250 0.34 0.167 3.25
400 0.0125 0.15 0.118 4.49
litz at 100 kHz: strands of 0.24 mm, about 128 strands for 5.6 mm2For a fixed inductor, raising the switching frequency reduces the ripple flux and therefore the core loss, while skin and proximity effects raise the AC resistance of solid wire. In an inductor, only the ripple part of the current sees the higher AC resistance, so solid wire is often acceptable when ripple is small; in transformers, where all the current is AC, litz or foil windings are essential at high frequency. The real design trade-off is that higher frequency allows a smaller inductor (with less copper and core), at the price of higher switching losses in the semiconductors (Lesson 1). The optimum frequency for silicon IGBT converters is typically 5 to 20 kHz, for silicon MOSFETs 50 to 200 kHz, and for SiC and GaN converters several hundred kilohertz to a few megahertz.
Transformers for switching converters
Isolated converters (Lesson 5) use transformers to provide galvanic isolation for safety and to match very different input and output voltages efficiently. A transformer is designed around volt-seconds: from Faraday's law, a square-wave voltage of amplitude \(V\) at frequency \(f\) applied to \(N_p\) turns produces a peak flux density:
so the minimum primary turns are \(N_p = V / (4 f B_{pk} A_e)\). Unlike an inductor, a transformer ideally stores no energy and has no gap; its magnetizing current, which sets up the flux, should be small. Two non-ideal elements matter: the magnetizing inductance, which carries the magnetizing current, and the leakage inductance, from flux that links one winding but not the other. Leakage inductance causes voltage spikes and ringing when switches turn off, which snubbers or clamps must absorb (Lesson 8), but resonant converters such as the LLC use it deliberately as part of their resonant tank. The flyback converter is a special case: its "transformer" is really a coupled inductor with a gap, storing energy while the switch is on and delivering it to the output when the switch is off.
Worked problem
Turns for a full-bridge transformer
A full-bridge converter applies a ±400 V square wave at 100 kHz to a transformer on a core with \(A_e = 353\) mm². The peak flux density is limited to 0.1 T to control core loss. The output is a 54 V bus, and the maximum effective duty cycle is 85 percent. Find the primary and secondary turns, neglecting rectifier drops.
Show solution
- Primary: \(N_p = 400 / (4 \times 100{,}000 \times 0.1 \times 353 \times 10^{-6}) = 400 / 14.1 = 28.3\), so 29 turns.
- Required ratio: the secondary must produce 54 V at 85 percent duty, so \(V_s = 54 / 0.85 = 63.5\) V; \(N_s = 29 \times 63.5 / 400 = 4.6\), so 5 turns.
- Check: with 5 turns, the secondary voltage is \(400 \times 5 / 29 = 69\) V, giving 54 V at a duty cycle of about 78 percent, within the limit.
Answer: A 29:5 turns ratio. The low flux density, much below saturation, is typical at 100 kHz, where core loss rather than saturation limits the design. With so few secondary turns carrying high current, the secondary is often made of copper foil or a printed circuit board winding.
Planar magnetics, with windings formed as copper layers in a printed circuit board and low-profile cores, are increasingly used in high-frequency converters such as the 48 V to processor-voltage converters in data center servers. They give repeatable leakage inductance, good cooling and low height, and allow matrix transformers that distribute current among several small cores.
Capacitors
Capacitors filter ripple, supply pulsed currents, hold up voltage during brief interruptions and form resonant circuits. Four families dominate power electronics:
- Aluminum electrolytic capacitors give the most capacitance per unit volume and cost, for DC links and bulk storage, but have relatively high ESR, limited ripple current and a life limited by electrolyte evaporation (Lesson 2).
- Polymer and hybrid aluminum capacitors use a solid or partly solid conductive polymer, giving lower ESR and longer life at lower voltages.
- Film capacitors (polypropylene) have low ESR, high ripple current capacity, self-healing behavior after small breakdowns and long life; they are preferred for DC links in drives, solar inverters and electric vehicle traction inverters, and for AC filters and snubbers.
- Multilayer ceramic capacitors (MLCCs) have very low ESR and inductance and are used for high-frequency decoupling and in the output filters of compact converters. Class II dielectrics such as X7R and X5R lose a large part of their capacitance under DC bias: a capacitor operated near its rated voltage may provide well under half its nominal value, so designers use the manufacturer's bias curves.
The ripple current a capacitor carries heats it through its ESR, and this heating, added to the ambient temperature, sets its core temperature and therefore its life. Selecting a DC link capacitor is usually a matter of meeting the ripple current rating, not the capacitance: a bank chosen for capacitance alone may overheat.
Worked problem
DC link capacitor bank
A drive's DC link carries 30 A RMS ripple current. Candidate electrolytic capacitors are rated at 4.5 A RMS ripple each at the expected temperature, with ESR of 25 mΩ. Film capacitors are rated at 20 A RMS each, with ESR of 3 mΩ. How many of each are needed, and what is the total ESR loss?
Show solution
- Electrolytic: \(30 / 4.5 = 6.7\), so 7 capacitors in parallel, each carrying \(30/7 = 4.3\) A; loss \(7 \times 4.3^2 \times 0.025 = 3.2\) W.
- Film: \(30 / 20 = 1.5\), so 2 capacitors, each carrying 15 A; loss \(2 \times 15^2 \times 0.003 = 1.35\) W.
Answer: Seven electrolytic or two film capacitors. The electrolytic bank provides far more capacitance, which may be needed for hold-up or for stability with the supply, while the film bank runs cooler and lasts longer. Many modern drives and inverters use film capacitors with less total capacitance and rely on control to handle the larger voltage ripple.
Common mistakes
MistakeUsing ferrite saturation flux density at room temperature.
InsteadUse the value at the highest operating temperature, about 0.35 T for many grades at 100 °C.
MistakeAssuming the gap formula gives exact inductance.
InsteadFringing raises inductance; split large gaps and measure the result.
MistakeUsing solid wire for high-frequency transformer windings.
InsteadUse litz wire, foil or interleaving to control skin and proximity losses.
MistakeChoosing MLCCs by nominal capacitance.
InsteadCheck DC bias derating curves for class II ceramics.
MistakeSizing a DC link capacitor bank by capacitance alone.
InsteadMeet the ripple current rating and check core temperature.
Summary and key terms
- Ampère's and Faraday's laws relate current, flux and voltage; reluctances add, and L = N²/ℛ.
- An air gap sets inductance, stores energy and prevents saturation under DC current.
- Core materials range from silicon steel and nanocrystalline alloys to powder cores and ferrites.
- The area product Ae·Aw ≥ L·Ipk·Irms/(Bmax·J·Ku) sizes an inductor core.
- Core loss follows the Steinmetz equation; skin and proximity effects raise winding loss at high frequency.
- Transformer primary turns follow from volt-seconds: Np = V/(4 f Bpk Ae); leakage inductance causes ringing.
- Capacitor selection is driven by ripple current, ESR heating, DC bias derating and life.
- Reluctance
- Opposition of a magnetic path to flux, l/(μA).
- Saturation
- The condition in which a core can carry little more flux and permeability collapses.
- Air gap
- A non-magnetic section in a core that controls inductance and stores energy.
- Area product
- The product of core area and window area, used to size a magnetic component.
- Window utilization
- The fraction of a core window filled with copper.
- Steinmetz equation
- An empirical fit of core loss to frequency and flux density.
- Skin depth
- The depth within which AC current flows near a conductor surface.
- Litz wire
- Wire made of many insulated, twisted strands to reduce high-frequency losses.
- Leakage inductance
- Inductance from flux linking one winding but not another.
- Equivalent series resistance (ESR)
- The resistance that causes a capacitor to heat with ripple current.
Practice problems
Work each problem on paper first, then reveal the answer to check.
A core has reluctance 2 × 10⁶ A/Wb. Inductance with 40 turns?
Show answer
1,600 / 2 × 10⁶ = 0.8 mH.
Why does an inductor's gap set its inductance?
Show answer
The gap's reluctance dominates the much smaller reluctance of the core.
Required area product for L = 50 µH, Ipk = 20 A, Irms = 18 A, Bmax = 0.3 T, J = 5 A/mm², Ku = 0.4?
Show answer
50e-6 × 20 × 18 / (0.3 × 5e6 × 0.4) = 3.0 × 10⁻⁸ m⁴ = 30,000 mm⁴.
Turns for 50 µH at 20 A peak, 0.3 T, Ae = 178 mm²?
Show answer
50e-6 × 20 / (0.3 × 178e-6) = 18.7, so 19 turns.
If α = 1.4 and β = 2.6, how does core loss change if frequency doubles at constant B?
Show answer
It rises by 2^1.4 ≈ 2.6 times.
Skin depth in copper at 400 kHz, given 0.24 mm at 100 kHz?
Show answer
Halved: about 0.12 mm.
Minimum primary turns for ±48 V at 200 kHz, Bpk 0.1 T, Ae 50 mm²?
Show answer
48 / (4 × 200,000 × 0.1 × 50e-6) = 12 turns.
Why is the flyback "transformer" gapped?
Show answer
It is a coupled inductor that stores energy each cycle.
Why might a 10 µF X7R capacitor give only 4 µF in a circuit?
Show answer
DC bias reduces the capacitance of class II ceramics.
A capacitor carries 6 A RMS with 20 mΩ ESR. Heat dissipated?
Show answer
0.72 W.
Lesson quiz · 5 questions
Score 60% or more to complete this lesson.
Create a free account to save your progress, take quizzes and earn a certificate.